A closed vessel A having volume V contains `N_(2)` at pressure P and temperature T. another closed vessel B having the same volume V contains. He at the same pressure P. but temperature 2T. The ratio of masses of `N_(2)` and He in the vesses A and B is

Options

A. `1:2`
B. `3:2`
C. `5:2`
D. `14:1`

Answer

To solve the problem, we need to find the ratio of the masses of nitrogen (N₂) in vessel A and helium (He) in vessel B. We will use the ideal gas law and the relationship between moles, mass, and molecular mass.

### Step-by-Step Solution:

1. **Identify Given Data:**
- For vessel A (N₂):
- Volume (V) = V
- Pressure (P) = P
- Temperature (T) = T
- For vessel B (He):
- Volume (V) = V
- Pressure (P) = P
- Temperature (T) = 2T

2. **Use the Ideal Gas Law:**
The ideal gas law is given by:
\[ PV = nRT \]
where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = universal gas constant
- \( T \) = temperature

3. **Calculate Moles for Each Gas:**
- For nitrogen (N₂) in vessel A:
\[ n_1 = \frac{PV}{RT} \]
Substituting \( T = T \):
\[ n_1 = \frac{PV}{RT} \]

- For helium (He) in vessel B:
\[ n_2 = \frac{PV}{R(2T)} \]
Substituting \( T = 2T \):
\[ n_2 = \frac{PV}{2RT} \]

4. **Relate the Moles:**
From the equations for \( n_1 \) and \( n_2 \):
\[ n_1 = \frac{PV}{RT} \quad \text{and} \quad n_2 = \frac{PV}{2RT} \]
We can express \( n_1 \) in terms of \( n_2 \):
\[ n_1 = 2n_2 \]

5. **Relate Mass to Moles:**
The mass of a gas can be expressed as:
\[ \text{mass} = n \times \text{molecular mass} \]
Therefore, for nitrogen (N₂):
\[ m_{N_2} = n_1 \times M_{N_2} \]
And for helium (He):
\[ m_{He} = n_2 \times M_{He} \]

6. **Substituting Moles into Mass Equations:**
Substitute \( n_1 \) and \( n_2 \):
\[ m_{N_2} = (2n_2) \times M_{N_2} \]
\[ m_{He} = n_2 \times M_{He} \]

7. **Calculate the Ratio of Masses:**
Now, we can find the ratio of the masses:
\[ \frac{m_{N_2}}{m_{He}} = \frac{(2n_2) \times M_{N_2}}{n_2 \times M_{He}} = \frac{2M_{N_2}}{M_{He}} \]

8. **Substituting Molecular Masses:**
The molecular mass of nitrogen \( M_{N_2} = 28 \, \text{g/mol} \) and for helium \( M_{He} = 4 \, \text{g/mol} \):
\[ \frac{m_{N_2}}{m_{He}} = \frac{2 \times 28}{4} = \frac{56}{4} = 14 \]

9. **Final Ratio:**
Thus, the ratio of the masses of nitrogen to helium is:
\[ \frac{m_{N_2}}{m_{He}} = 14:1 \]

### Conclusion:
The ratio of the masses of \( N_2 \) and He in the vessels A and B is \( 14:1 \).